be real numbers less than 2.Then prove that 
be the sequence of all natural numbers which are sum of squares of two natural numbers.
which 
.
intersect at 
,
tangent to the extensions of 
at 
respectively. Circle 
passes through points 
,
.
.
such that 
for reals 
.
Y by might_guy, Adventure10, Mango247, and 8 other users
Geometry is my weakest part although i have tried very hard to improve lately. I want to ask geometry experts of this forum : How do you train yourself on geometry ? What is your way of thinking when dealing with a hard geometry problem like 2008/6 or 2011/6 ? In my country, during national olimpiad and tst the geometry problems are usually easy or medium problems and i feel like we are not trained well enough. When i tried 2008/6 or 2011/6 i got stuck with no idea at all. Can you share what's your way of cracking it ? Did you deal with other problems with similar kind of diagram before so you can relate to it? (like in 2011/6 the fact that the the incenter of the determined triangle lies on the circumcircle of triangle ABC appears in previous test). Or did you try to draw a very good diagram and try to guess some properties ?
Also what books or problem sets do you think is appropriate for training ?
Sorry for this long post and my poor english
Also what books or problem sets do you think is appropriate for training ?
Sorry for this long post and my poor english
This post has been edited 2 times. Last edited by inge, Nov 27, 2020, 8:00 AM
(which amusingly came up as USA JMO 2014 #6) then the solution path is quite natural. If not, you are at a severe disadvantage. (Shameless plug: I'm in the process of trying to get a geometry book published (
,
.
and 
are. The diagram gives you an immediate, rapid way to disprove false conjectures, and also suggests new ones.
be the tangency point of the two circles, and 
be the triangle formed by 
,
,
.
,
,
be the second intersections onto 
.
and 
look true as well. (I also discovered and proved the incenter lemma from my diagram, but I did not end up using it in my solution. However, it would certainly have been worth marks at the IMO.)![[asy]
size(8cm);
defaultpen(fontsize(8pt));
pair A = dir(110);
dot("$A$", A, dir(A));
pair B = dir(195);
dot("$B$", B, dir(160));
pair C = dir(325);
dot("$C$", C, 1.4*dir(30));
pair P = dir(270);
dot("$P$", P, dir(P));
draw(unitcircle);
draw(A--B--C--cycle, blue);
pair U = P+(2,0);
pair V = 2*P-U;
pair X_1 = reflect(B,C)*P;
pair Y_1 = reflect(C,A)*P;
pair Z_1 = reflect(A,B)*P;
pair X_2 = extension(B, C, U, V);
dot(X_2);
pair Y_2 = extension(C, A, U, V);
dot(Y_2);
pair Z_2 = extension(A, B, U, V);
dot(Z_2);
draw(B--Z_2, dotted+blue);
draw(C--Y_2, dotted+blue);
draw(C--X_2, dotted+blue);
draw(X_2--Z_2);
pair A_1 = extension(Y_1, Y_2, Z_1, Z_2);
dot("$A_1$", A_1, dir(A_1));
pair B_1 = extension(Z_1, Z_2, X_1, X_2);
dot("$B_1$", B_1, dir(B_1));
pair C_1 = extension(X_1, X_2, Y_1, Y_2);
dot("$C_1$", C_1, dir(50));
draw(A_1--B_1--C_1--cycle, green);
draw(C_1--X_2, dotted+green);
draw(circumcircle(A_1, B_1, C_1));
pair A_2 = A*A/P;
dot("$A_2$", A_2, dir(-20));
pair B_2 = B*B/P;
dot("$B_2$", B_2, dir(130));
pair C_2 = C*C/P;
dot("$C_2$", C_2, dir(C_2));
draw(A_2--B_2--C_2--cycle, red);
pair T = extension(A_1, A_2, B_1, B_2);
dot("$T$", T, dir(T));
draw(T--A_1, dashed);
draw(T--B_1, dashed);
draw(T--C_1, dashed);
[/asy]](http://latex.artofproblemsolving.com/8/5/c/85cee610e59798effe2c659c363c0f92f7870415.png)
defined are homothetic. So my conjecture is almost definitely true, and I simply have to show that 
,
,
concur on 
,
as the reference triangle, it is fairly straightforward to compute the coordinates of 
,
,
.
.
,
lying on the incircle.
